What is the Kronecker product of two 1D vectors?

Question

When working with $n$-qubit systems, we can use the Kronecker product $\otimes$ to build corresponding bigger Hilbert spaces. For example:

$$\begin{bmatrix} \alpha \\ \beta \end{bmatrix} \otimes \begin{bmatrix} \gamma \\ \delta \end{bmatrix} = \begin{bmatrix} \alpha\gamma \\ \alpha \delta \\ \beta\gamma \\ \beta \delta \end{bmatrix},$$

and we can keep building up from that. However, I was wondering if it is possible to write $\begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ as a Kronecker product itself? Intuitively I'd write something like $\begin{bmatrix} \alpha \end{bmatrix} \otimes \begin{bmatrix} \beta \end{bmatrix}$, although if I use the formal definition then the result would be $\begin{bmatrix} \alpha \beta \end{bmatrix}$, which is still a $1\times1$ matrix and not what I expect. I can hand-wavily say that when when I'm building a bigger Hilbert space out of $\begin{bmatrix} \alpha \end{bmatrix}$ and $\begin{bmatrix} \beta \end{bmatrix}$, they each need to go on a different dimension because each one lives on its own Hilbert space, but how do I justify that mathematically?

Answer 1

The tensor product of $\mathbb C^m$ and $\mathbb C^n$ is (isomorphic to) $\mathbb C^{mn}$. Therefore, if you're asking the reverse question it becomes a matter of factoring numbers. In your case your target space is $\mathbb C^2$ and there are, obviously, only two ways to factor the number two: $2\cdot 1$ and $1\cdot 2$. Therefore, the only ways to build your target space are: $$ \mathbb C^2=\mathbb C^2\otimes\mathbb C^1\quad\text{ and }\quad\mathbb C^2=\mathbb C^1\otimes\mathbb C^2 $$ In particular, there is no non-trivial way of decomposing $\mathbb C^2$ because $2$ is a prime number.

Answer 2

It is possible to do this, but not in a nontrivial way: \begin{align*} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} &= \begin{bmatrix} \alpha \\ \beta \end{bmatrix}\otimes \begin{bmatrix}1\end{bmatrix} \end{align*} This agrees with the answer of @frederik-vom-ende.